A diode has the I–V characteristic shown, with knee voltage 0.8 V. It feeds the initially discharged 10 μF capacitor from $V_{in}=100\sin(1000t)$ V. Reverse breakdown greatly exceeds 100 V. Find steady-state capacitor voltage Vout in volts.
During positive peaks the diode charges the capacitor until its voltage is the peak input minus the forward drop: 100−0.8=99.2 V. There is no discharge load, and the diode blocks the reverse half-cycle, so the capacitor retains this DC peak-detector voltage. Thus A.
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